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Let's say we are working with the field extension $\mathbb{Q}(\gamma)$, where $\gamma$ is the seventh root of unity. I know my basis for this extension will thus be:

$\{1, \gamma, \gamma^2, \gamma^3, \gamma^4, \gamma^5, \gamma^6 \}$

And the Galois group will consist of automorphisms that take the generator $\gamma$ and map it to it's different powers. Since this group has order $6$, I know I will have a subgroup of order $2$ and a subgroup of order$3$. The subgroup of order $2$ will be:

$<\sigma_2>$ where $\sigma_2$ sends $\gamma$ to $\gamma^2$

And the subgroup of order 3 will be:

$<\sigma_6>$ where $\sigma_6$ sends $\gamma$ to $\gamma^6$

My goal is to find the subfield that is fixed by, say, the subgroup $<\sigma_2>$

What I did was take an arbitrary element of $\mathbb{Q}(\gamma)$ and acted on it with $\sigma_2$. The arbitrary element $x$ will just be a linear combination of the basis elements of $\mathbb{Q}(\gamma)$, thus:

$x = a + b\gamma + c\gamma^2 + d\gamma^3 + e\gamma^4 + d\gamma^5 + e\gamma^6$

Then:

$\sigma_2(x)$ = $\sigma_2(a) + \sigma_2(b\gamma) + \sigma_2(c\gamma^2) + \sigma_2(d\gamma^3) + \sigma_2(e\gamma^4) + \sigma_2(d\gamma^5) + \sigma_2(e\gamma^6) = a + b\gamma^2 + c\gamma^4 + d\gamma^6 + e\gamma + f\gamma^3 + g\gamma^5$

Comparing coefficients, we see that $a = a$ and $b = e, c = b, d = f, e = c, d = g, e = d$

When we did a similar exercise in class, I remember things simplified much more nicely. From the looks of it, my calculation shows that things get shuffled around, so the only things fixed are the coefficients from $\mathbb{Q}$, but I know that's not right. What am I doing wrong?

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3 Answers 3

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You've got the orders wrong, $\sigma_2$ has order $3$ and $\sigma_6$ has order $2$.

Clearly the fixed field of $\zeta_2$ is the unique quadratic subfield (since the index of the subgroup is $2$) which we know to be $\Bbb Q(\sqrt{-7})$ by Quadratic Reciprocity since $7\equiv 3\mod 4$. The real subfield is implicitly $K\cap\Bbb R$, and explicitly $\Bbb Q(\zeta_7+\zeta_7^{-1})$ since this is clearly fixed by $\sigma_6$ but not $\sigma_2$.

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Your above writings imply that $\mathbb{Q}(\zeta)$ has degree $7$ over $\mathbb{Q}$. What is wrong?

Try writing $\zeta^{6}$ in terms of $1,\dots,\zeta^{5}$. Remember the minimal polynomial for $\zeta$.

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Just to suggest a different approach, instead of seeing what an automorphism does on a general element, try looking at the sum of the orbit on your generator $\zeta_7$. So for instance, with your $\sigma_2$, we have $\sigma_2(\zeta)=\zeta^2$, $\sigma_2(\zeta^2)=\zeta^4$, and $\sigma_2(\zeta^4)=\zeta$, so this is the full orbit, so look at $\zeta+\zeta^2+\zeta^4$. It is clearly fixed by $\zeta_2$, so is contained in the fixed field. You just want to show it generates, so must have degree 2. To do this, you need to take powers and see if you can find a degree 2 polynomial it satisfies.

The other one is similar: the orbit will be $\zeta+\zeta^6$, and you can show that is degree 3 either by finding the minimal polynomial or by observing this is $2\cos(2\pi/7)$ using De Moivre and this is real, and try to show $\zeta$ lies in a degree 2 extension of this field, so the subfield must be degree 3 by the Tower Law and is clearly fixed by the other subgroup.

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